$f(x, y) = (6y^2x^3 - 2x^2, xy^3)$ What is the curl of $f$ at $(1, 3)$ ?
Solution: The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ xy^3 \right] \\ \\ &= y^3 \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ 6y^2x^3 - 2x^2 \right] \\ \\ &= 12yx^3 \end{aligned}$ Therefore: $\text{curl}(f) = y^3 - 12yx^3$ The curl of $f$ at $(1, 3)$ is $-9$.